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proof of existence of the essential supremum
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(Proof)
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Suppose that $(\Omega,\mathcal{F},\mu)$ is a $\sigma$ -finite measure space and $\mathcal{S}$ is a collection of measurable functions $f\colon\Omega\rightarrow\mathbb{\bar R}$ . We show that the essential supremum of $\mathcal{S}$ exists and furthermore, if it is nonempty then there is a sequence $f_n\in\mathcal{S}$ such that \begin{equation*} \esssup\,\mathcal{S}=\sup_nf_n. \end{equation*} As any $\sigma$ -finite measure is equivalent to a probability measure, we may suppose without loss of generality that $\mu$ is a probability measure. Also, without loss of generality, suppose that $\mathcal{S}$ is nonempty, and let $\mathcal{S}^\prime$ consist of the collection of maximums of finite sequences of functions in $\mathcal{S}$ . Then choose any continuous and strictly increasing $\theta\colon\mathbb{\bar R}\rightarrow\mathbb{R}$ . For example, we can take
As $\theta(f)$ is a bounded and measurable function for all $f\in\mathcal{S}^\prime$ , we can set \begin{equation*} \alpha=\sup\left\{\int\theta(f)\,d\mu:f\in\mathcal{S}^\prime\right\}. \end{equation*}Then choose a sequence $g_n$ in $\mathcal{S}^\prime$ such that $\int\theta(g_n)\,d\mu\rightarrow\alpha$ . By replacing $g_n$ by the maximum of $g_1,\ldots,g_n$ if necessary, we may assume that $g_{n+1}\ge g_n$ for each $n$ . Set \begin{equation*} f=\sup_n g_n. \end{equation*}Also, every $g_n$ is the maximum of a finite sequence of functions $g_{n,1},\ldots,g_{n,{m_n}}$ in $\mathcal{S}$ . Therefore, there exists a sequence $f_n\in\mathcal{S}$ such that \begin{equation*} \{f_1,f_2,\ldots\}=\{g_{n,m}:n\ge 1,1\le m\le m_n\}. \end{equation*}Then, \begin{equation*} f=\sup_nf_n. \end{equation*}It only remains to be shown that $f$ is indeed the essential supremum of $\mathcal{S}$ . First, by continuity of $\theta$ and the dominated convergence theorem, \begin{equation*} \int\theta(f)\,d\mu=\lim_{n\rightarrow\infty}\int\theta(g_n)\,d\mu=\alpha. \end{equation*}Similarly, for any $g\in\mathcal{S}$ , \begin{equation*} \int\theta(f\vee
g)=\lim_{n\rightarrow\infty}\int\theta(g_n\vee g)\,d\mu\le\alpha. \end{equation*}It follows that $\theta(f\vee g)-\theta(f)$ is a nonnegative function with nonpositive integral, and so is equal to zero $\mu$ -almost everywhere. As $\theta$ is strictly increasing, $f\vee g=f$ and therefore $f\ge g$ $\mu$ -almost everywhere.
Finally, suppose that $g\colon \Omega\rightarrow\mathbb{\bar R}$ satisfies $g\ge h$ ($\mu$ -a.e.) for all $h\in\mathcal{S}$ . Then, $g\ge f_n$ and, \begin{equation*} g\ge\sup_nf_n=f \end{equation*}$\mu$ -a.e., as required.
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"proof of existence of the essential supremum" is owned by gel.
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| Keywords: |
measure space, essential supremum,  -finite |
This object's parent.
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Cross-references: integral, dominated convergence theorem, bounded, strictly increasing, continuous, functions, finite sequences, probability measure, without loss of generality, sequence, essential supremum, measurable functions, collection, measure space
This is version 4 of proof of existence of the essential supremum, born on 2008-12-27, modified 2009-05-30.
Object id is 11400, canonical name is ProofOfExistenceOfTheEssentialSupremum.
Accessed 491 times total.
Classification:
| AMS MSC: | 28A20 (Measure and integration :: Classical measure theory :: Measurable and nonmeasurable functions, sequences of measurable functions, modes of convergence) |
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Pending Errata and Addenda
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